# Dense and nowhere dense

Let $X$ be a topological space and $A$ be a non-empty subset of $X$. Then one can conclude that if $X\setminus A$ is nowhere dense in $X$, $A$ is dense in $X$,

Is the above statement true in general? I know if $A$ is open then the result is true, but I am not sure otherwise.

No, that is not true in general.

Consider $X = \mathbb{R}$. Clearly $A = \mathbb{Q}$ is dense in $\mathbb{R}$, but $\mathbb{R} \setminus \mathbb{Q}$ is also known to be dense as well.

• Thank you so much – Golam biswas Jun 25 '18 at 17:40

No, as others pointed out already. Sets can be dense and co-dense, like the rationals and the irrationals in the real numbers.

But $A$ is nowhere dense iff $X\setminus \overline{A}$ is dense. So something close to it is true.

$$A$$ is dense $$\Rightarrow A^-=X\Rightarrow A^e=\emptyset \Rightarrow A^{c\circ}=\emptyset$$. $$A^{c\circ}=\emptyset$$ does not always imply $$A^{c-\circ}=\emptyset$$, since $$A^c\subseteq A^{c-}$$.